Question

A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of $400pm$. If the density of the substance in the crystal is $8gc{m}^{-3}$ , then the number of atoms present in $128g$ of the crystal is

• A. $2\times 10^{24}$
• B. $4\times 10^{24}$
• C. $1\times 10^{24}$
• D. $3\times 10^{24}$

Answer 1

Answer: (C)

$d=\frac{Z\times M}{a^3\times N_A}\Rightarrow M=\frac{a^3\times d\times N_A}{Z}$
$M=\frac{(400{)}^3\times 10^{-30}\times 8\times N_A}{4}$
On solving $M=64\times 2\times 10^{-24}$ $N_{A}g$
Now, $\left(64\times 2\times 10^{-24}{N}_A\right)$ g contains $N_A$ atoms
$128g$ will contain $=\frac{N_A}{64\times 2\times 10^{-24}{N}_A}\times 128$
$=1\times 10^{24}$